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The title says it all:

Let $A$ be a commutative ring.

Are there any interesting known conditions on $A$ (other then being noetherian of course...) to ensure existence of a (non-zero) commutative noetherian ring $B$, and a flat ring map $A\to B$?

the L
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2 Answers2

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If $A$ is integral, you can take $B$ to be the fraction field $K(A)$ of $A$. The morphism $A\to K(A)$ is flat. In general, you (edit: can't) probably take $B$ to be the total field of fractions.

Harry
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  • Are you claiming that the generalized fraction field will always be noetherian? – the L Nov 13 '12 at 22:12
  • Actually, let me take that back. That can't be correct. What about taking $B$ to be the localization of $A$ at some minimal prime ideal? In this case $B$ is a local zero-dimensional ring. Aren't these always noetherian? – Harry Nov 13 '12 at 22:19
  • I proved a theorem which is correct for any ring $A$ which satisfies the condition of the above question. I know that my theorem is not true for any ring, so assuming my proof is correct, the answer to the above question cannot be - for any ring. – the L Nov 13 '12 at 22:20
  • Dear @Harry: Dimension zero local rings do not have to be Noetherian. See the link in the comments to the answer below. – Rankeya Nov 14 '12 at 00:29
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Inspired by Harry's observation in the discussion, here is another way to get examples of what you seek. If you take any ring $A$ having a minimal prime ideal $p$ that is finitely generated (but $A$ is not necessarily Noetherian), then $A_p$ is a Noetherian ring (this follows from the result that if all the prime ideals of a ring are finitely generated, then the ring is Noetherian). You will then get a flat ring map $A \rightarrow A_p$, but $A$ is not Noetherian.

Rankeya
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  • Thanks, so am I correct that not all rings have minimal primes which are finitely generated? – the L Nov 13 '12 at 22:32
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    Yes you are. See the discussion here: http://math.stackexchange.com/questions/39458/non-noetherian-ring-with-a-single-prime-ideal – Rankeya Nov 13 '12 at 22:38
  • Thanks for both of you. That's funny, because now I actually proved that there is a ring which does not satisfy the above question. Any flat extension of it is not noetherian :) – the L Nov 13 '12 at 22:57
  • well, any domain will do the case... – the L Nov 13 '12 at 23:40
  • You are correct. I will edit my answer. – Rankeya Nov 14 '12 at 00:13
  • For an example of a non-Noetherian ring of dim $\geq 1$, I think you can take the ring $k[x_1, x_2, \dots]/(x^2_1)$, and the minimal prime ideal $(x_1)$. – Rankeya Nov 14 '12 at 00:20
  • Also, I am not sure why you say dimension $0$ rings have only one prime ideal. – Rankeya Nov 14 '12 at 00:22
  • @anonymous: so what is the example of a ring without flat noetherian extension ? –  Nov 14 '12 at 17:38
  • There was an unfortunate mistake in my proof, so I don't know it anymore :( – the L Nov 14 '12 at 22:55